Question

Calculate the Q-value in the following decay:
${}_8^{19}O{\to }_9^{19}F+e^{-}+\overline{v}.$
These may look like long calculations; use the unit amu or u for the masses, and $c^2=931\frac{MeV}{u},$ and it will become much simpler.
$Atomicmasses{:}_8^{19}O:19.003576u;{}_9^{19}F:18.998403u.$

• A. 2.021 MeV
• B. 3.754 MeV
• C. 4.816 MeV
• D. 5.165 MeV

Answer 1

The correct option is C 4.816 MeV

The Q-value is the energy released in a radioactive process, and is equal to $E_Q=\Delta m{c}^2,$ where $\Delta m$ is the mass defect. Since the antineutrino $\overline{v}$ has almost zero mass, we will compare masses of the nuclei and $e^{-}$ only
${}_8^{19}O{\to }_9^{19}F+e^{-}+\overline{v}.$
What we are given are the atomic masses, which include all the electrons in the atoms present. In this calculation, we will denote atomic mass of X as m(X) and nuclear mass as $m_n\left(X\right),$ okay? The Q-value can be calculated as follows:
Q_value=$\Delta m{c}^2$
=(Initial mass-energy)-(Final mass-energy)
$=m_n{(}_8^{19}O)-[m_n{(}_9^{19}F)+m(e^{-}\left)\right]c^2$
$=m_n{(}_8^{19}O)-m_n{(}_9^{19}F)-[9\times m(e^{-})-8\times m(e^{-}\left)\right]c^2$
$=\left[m_n{(}_8^{19}O\right)+8\times m\left(e^{-}\right)]-[m_n{(}_9^{19}F)+9\times m(e^{-}\left)\right]c^2$
$=\left[m{(}_8^{19}O\right)-m{(}_9^{19}F\left)\right]c^2$
Where $m{(}_8^{19}O\left)andm{(}_9^{19}F\right)$ are the atomic masses, not nuclear masses anymore. Pretty smart, right? You don't have to derive this if you just remember that in a $\beta$ decay, the Q-value is also equal to the difference in the atomic masses multiplied by $c^2.$
Therefore, the Q-value of this process will be
$=\left[m{(}_8^{19}O\right)-m{(}_9^{19}F\left)\right]\times c^2$
$=[19.003576u-18.998403u]\times 931\frac{MeV}{u}$
$=\begin{array}{c}4.816MeV\end{array}.$